php - Ajax /PHP : Returning values

Question

问:我找不到我的代码不起作用的原因。

index.php

while($row = mysqli_fetch_array($result))
{

$Tri_CAPTION=$row['BOOKCAPTION'];
$Tri_IMAGE=$row['BOOKIMAGE'];
$BOOKT=$row['BOOKTITLE'];
$BOOKID=$row['BOOKID'];
$html = $Tri_IMAGE;
$doc = new DOMDocument();
$doc->loadHTML($html);
$xpath = new DOMXPath($doc);
$src = $xpath->evaluate("string(//img/@src)"); # "/images/image.jpg"


echo "<div class=\"mybooks\">
 <a  href=\"".$src."\" data-title=\"$BOOKT\"  data-lightbox=\"example-3\"  alt=\"Home\">
 <img src=\"".$src."\" width=\"150\" height=\"150\"/>  
</a>
 <br />
Title:<strong> 
".$BOOKT."</strong>
<br />

<a onClick=\"forusedata($BOOKID);\" href=\"#\">View Book</a> 

</div>";
  }
?>

Ajax :

<script>    
  function forusedata(newid)
  {         

$.post('insert_home.php',{newid:newid}).done(function(data){
    alert(data);

    });
        $('#Page1').dialog('open');
  }

</script>

insert_home.php

else if(isset($_POST['newid']))
{
$newid = ($_POST['newid']);

$result = mysqli_query($con,"SELECT BOOKCAPTION FROM book WHERE BOOKID='$newid'");
while($row = mysqli_fetch_array($result)){
echo $row['BOOKCAPTION'];
}
 mysqli_close($con);
}

Answer 1

为什么你的 insert_home.php 代码以 else if 开头？上面是不是少了什么东西？由于您没有说出实际上不起作用的内容，因此我猜测您没有收到任何警报数据。

我会将你的 insert_home.php 更改为:

if(isset($_POST['newid'])) {
    $newid = ($_POST['newid']);
    $result = mysqli_query($con,"SELECT BOOKCAPTION FROM book WHERE BOOKID='$newid'");
    while($row = mysqli_fetch_array($result)){
        echo $row['BOOKCAPTION'];
    }
    mysqli_close($con);
}
else {
    echo ('no results');
}

这会让您知道问题是出在 ajax 还是出在 insert_home.php 页面。